Given:
Iron, 125 grams
T
1 = 23.5 degrees Celsius, T2 =
78 degrees Celsius.
Required:
Heat produced in kilojoules
Solution:
The molar mass of iron is 55.8
grams per mole. SO we need to change the given mass of iron into moles.
Number of moles of iron = 125 g/(55.8
g/mol) = 2.24 moles
<span>
Q (heat) = nRT = nR(T2 = T1)</span>
Q (heat) = 2.24 moles (8.314
Joules per mol degrees Celsius) (78.0 degrees Celsius – 23.5 degrees Celsius)
<u>Q (heat) = 1014.97 Joules or
1.015 kilojoules</u>
<span>This is the amount of heat
produced in warming 125 g f iron.</span>
CO2 ; H20- They are the only ones that, on both sides, combined with another element and bonding of atoms
Answer: it is 5.5 mg
Explanation:
you have to multiply the mass value by 1000
Answer:
Kc = 2.34 mol*L
Explanation:
The calculation of the Kc of a reaction is performed using the values of the concentrations of the participants in the equilibrium.
A + B ⇄ C + D
Kc = [C] * [D] / [A] * [B]
According to the reaction
Kc = [SO2]^2 * [O2]^2 / [SO3]^2
Knowing the 0.900 mol of SO3 is placed in a 2.00-L it means we have a 0.450 mol/L of SO3
0.450 --> 0 + 0 (Beginning of the reaction)
0.260 --> 0.260 + 0.130 (During the reaction)
0.190 --> 0.260 + 0.130 (Equilibrium of the reaction)
Kc = [0.260]^2 + [0.130]^2 / [0.190]^2
Kc = 2.34 mol*L
